You are pushing a refrigerator across the floor of your kitchen. You exert a horizontal force of 294 N for 7.6 s, during which time the refrigerator moves a distance of 3.2 m at constant velocity.
(a) What is the total work (by all forces) done on the refrigerator?
(b) What is the work done by friction?
work=294*3.2
friction=above work done, since there is no change in KE, or PE
To find the answers to these questions, we need to understand the concept of work done and its relation to force and distance.
Work is defined as the product of the force applied on an object and the displacement of the object in the direction of the force. Mathematically, work (W) can be calculated using the following formula:
W = F * d * cos(theta),
where F is the force applied, d is the displacement, and theta is the angle between the force and displacement vectors.
Now, let's solve the given problem.
(a) What is the total work (by all forces) done on the refrigerator?
In this case, the refrigerator is moving at a constant velocity, which means there is no acceleration. This implies that the net force acting on the refrigerator is zero. Therefore, the total work done by all forces is also zero.
(b) What is the work done by friction?
Since the refrigerator is moving at a constant velocity, the work done by friction is equal in magnitude but opposite in direction to the work done by the applied force.
The work done by friction can be calculated using the same formula: W = F * d * cos(theta). Here, the force of friction (F_friction) is equal to the applied force (F_applied) because the refrigerator is moving at a constant velocity.
So, the work done by friction is:
W_friction = F_friction * d * cos(theta).
Now, let's substitute the given values into the formula to find the answer.
F_applied = 294 N,
d = 3.2 m,
theta = 180 degrees (the angle between the applied force and the displacement is 180 degrees, as they are in opposite directions).
W_friction = F_friction * d * cos(theta) = (294 N) * (3.2 m) * cos(180 degrees) = (-294 N) * (3.2 m) * (-1) = 940.8 J.
Therefore, the work done by friction on the refrigerator is 940.8 Joules.